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  "body": "\n## Proof\n\n1. Let $(P, \\le)$ be a partially ordered set, let $S \\subseteq P$, and let $b$ and $b'$ be [[suprema|0BTMQ04556]] of $S$.\n2. By the first condition of the definition, $b$ is an [[upper bound|0B5H2V4JW8]] of $S$.\n3. By the second condition for $b'$, applied to the upper bound $b$, we get $b' \\le b$.\n4. In the same way, $b'$ is an upper bound of $S$ and $b$ is a supremum, so $b \\le b'$.\n5. The relation $\\le$ is a [[partial order|011C29Q1S1]], so it is antisymmetric. From $b' \\le b$ and $b \\le b'$ we get $b = b'$.\n\n## Idea\n\nEach supremum is an upper bound, and each supremum lies below every upper bound. So two suprema lie below each other, and antisymmetry makes them equal.\n",
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